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Exercise 7.1.7

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1.

U^{k} (x) = 0

, then

U^{k + 1} (x) = U^{k} (U (x)) = 0

2.

We know

U^{m + 1} (V) = U^{m} (U (V)) \subset U^{m} (V)

. With the assumption

rank (U^{m}) = rank (U^{m + 1})

, we know that

U^{m + 1} (V) = U^{m} (V) .

This means

U (U^{m} (V)) = U^{m} (V)

and so $U^{k} (V) = U^{m} (V)$ for all integer $k \geq m$ .

3.

The assumption

rank (U^{m}) = rank (U^{m + 1})

implies

null (U^{m}) = null (U^{m + 1})

by Dimension Theorem. This means $N (U^{m}) = N (U^{m + 1})$ by the previous argument. If $U^{m + 2} (x) = 0$ , then $U (x)$ is an element in $N (U^{m + 1}) = N (U^{m})$ . Hence we have $U^{m} (U (x)) = 0$ and thus $x$ is an element in $N (U^{m + 1})$ . This means that $N (U^{m + 2}) \subset N (U^{m + 1})$ and so they are actually the same. Doing this inductively, we know that $N (U^{m}) = N (U^{k})$ for all integer $k \geq m$ .

4.

By the definition of

K_{λ}

, we know

K_{λ} = ∪_{p \geq 1} N ({(T - λI)}^{p}) .

But by the previous argument we know that

N ({(T - λI)}^{m}) = N ({(T - λI)}^{k})

for all integer $k \geq m$ and the set is increasing as $k$ increases. So actually $K_{λ}$ is $N ({(T - λI)}^{m})$ .

5.

Since the characteristic polynomial splits, the transformation

T

is diagonalizable if and only if

K_{λ} = E_{λ}

. By the previous argument, we know that

K_{λ} = N (T - λI) = E_{λ} .

6.

λ

is an eigenvalue of

T_{W}

, then

λ

is also an eigenvalue of

T

by Theorem 5.21. Since

T

is diagonalizable, we have the condition

rank (T - λI) = rank ({(T - λI)}^{2})

and so

N (T - λI) = N ({(T - λI)}^{2})

by the previous arguments. This implies

N (T_{W} - λ I_{W}) = N ({(T_{W} - λ I_{W})}^{2}) .

By Dimension Theorem, we get that

rank (T_{W} - λ I_{W}) = rank ({(T_{W} - λ I_{W})}^{2}) .

So $T_{W}$ is diagonalizable.

jephianlin

2011-06-27 00:00

Exercise 7.1.7

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